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1 year ago

A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is

Physics

1 answer:

Radda [10]1 year ago

8 0

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

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The strength of intermolecular forces between particles affects physical properties of substances such as boiling point, melting

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The right answer is h2o

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2 years ago

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A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

Linear mass density= 0.0127 kg/m

Velocity= $2*\frac{l}{t}$

Velocity= 2 * $\frac{65*10^-2}{7.84}$

Velocity= 165.8 m/s

So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

165.8= $\sqrt{\frac{m*9.81}{0.0127} }$

Solving we get

m= 35.58 kg

or m= 35.6 kg

3 0

2 years ago

you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre

Sidana [21]

Answer:

Explanation:

Let the height of the tree is y and the distance of tree from point B is x.

According to the diagram

$tan61 = \frac{y}{x}$

x = 0.55 y ..... (1)

$tan49.5 = \frac{y}{50+x}$

(50 + 0.55y) 1.17 = y ..... from equation (1)

58.5 + 0.644 y = y

0.356 y = 58.5

y = 164.3 ft

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2 years ago

An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on

marysya [2.9K]

Answer:

The answer is: c. It does not move

Explanation:

Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.

6 0

1 year ago

Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m

Nimfa-mama [501]

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

= 0.5 m

7 0

2 years ago